TITLE: 4-coloring maps of pentagons
QUESTION [6 upvotes]: Is there a simple proof for the 4-color theorem when restricted to (finite) maps all of whose
(internal) regions are pentagons?  I am in fact most interested in convex pentagons, if that additional
structure helps.  These maps lead to graphs of maximum degree $\Delta=5$, so a result for
such graphs will answer my question.
If all regions are quadrilaterals, then four colors are sometimes necessary,
even if all quadrilaterals are convex, e.g.,


But here there is a simple proof of 4-colorability:
Identify a quadrilateral with an exposed edge, remove it, 4-color the remainder
by induction, and replace the quad, coloring it with a color different from its
at most three neighbors.
Obviously this simple proof fails for pentagons.
Especially the $\Delta=5$ case is likely known to the experts.  So a reference would suffice.
Thanks!

REPLY [10 votes]: If $v$ is a vertex of degree at most 4 in a planar graph $G$ then one can extend a proper 4-coloring of $V(G) \setminus \{v\}$ to $v$ after possibly modifying it using the classical Kempe chain argument. See for example paragraph 5 of the "Summary of proof ideas" section of the Wikipedia entry on the 4-color theorem: http://en.wikipedia.org/wiki/Four_color_theorem
As pentagons with exposed edges correspond to vertices of degree 4 in the dual graph, one can color the map by induction using this trick.